Khalil Jones

2023-03-14

The resistance of a wire of iron is 10 ohms and temp. coefficient of resistivity is 5 \times 10^{−3}/^{\circ} C. At 20^{\circ} C it carries 30 milliamperes of current. Keeping constant potential difference between its ends, the temperature of the wire is raised to 120^{\circ} C. The current in milliamperes that flows in the wire is

Veronella6nk

Step 1: Calculation of resistance at
Let variables at be marked by subscript 1 and at be marked by subscript 2.
Given that,
$\frac{{R}_{1}}{{R}_{2}}=\frac{1+\alpha {t}_{1}}{1+\alpha {t}_{2}}$
Where ${t}_{1}={20}^{\circ }C$ and ${R}_{1}$ is resistance at ${t}_{1}$ and ${t}_{2}={120}^{\circ }C$ and ${R}_{2}$ is resistance at ${t}_{2}$.
We have,
${R}_{1}=10\mathrm{\Omega },\alpha =5×{10}^{-3}{/}^{\circ }C\phantom{\rule{0ex}{0ex}}\therefore \frac{10}{{R}_{2}}=\frac{1+\left(5×{10}^{-3}×20\right)}{1+\left(5×{10}^{-3}×120\right)}\phantom{\rule{0ex}{0ex}}⇒{R}_{2}\approx 15\mathrm{\Omega }$
Step 2: Calculation of current at ${120}^{\circ }C$
when the potential difference is constant,
$i\propto \frac{1}{R}\phantom{\rule{0ex}{0ex}}\therefore \frac{{i}_{1}}{{i}_{2}}=\frac{{R}_{2}}{{R}_{1}}=\frac{15}{10}=1.5$
It is also provided,
$i1=30mA\phantom{\rule{0ex}{0ex}}\therefore i2=301.5\phantom{\rule{0ex}{0ex}}⇒i2=20mA$
Therefore, current flowing through the circuit at ${120}^{\circ }$ C is 20mA.

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